Question

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.)

Answer 1

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling:

`\displaystyle(\sigma)/(√(n)) = (1)/(\sqrt4) = 0.5`

We have to find the value of x such that the probability is 0.05

P(X > x) = 0.05

`P( X > x) = P( z > \displaystyle(x - 40)/(0.5))=0.05`

`= 1 -P( z \leq \displaystyle(x - 40)/(0.5))=0.05`

`=P( z \leq \displaystyle(x - 40)/(0.5))=0.95`

Calculation the value from standard normal z table, we have,

`\displaystyle(x - 40)/(0.5) = 1.64\\x = 40.825 \approx 40.83`

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.